Henry Bottomley

5 Leydon Close, London SE16 5PF, United Kingdom

E-mail:
SE16@btinternet.com

This note considers eleven particular families of interrelated
multiplicative functions, many of which are listed in Smarandache's problems.

These are multiplicative in the sense that a function f(n) has the property
that for any two coprime positive integers a and b, i.e. with a highest common
factor (also known as greatest common divisor) of 1, then f(a*b)=f(a)*f(b). It
immediately follows that f(1)=1 unless all other values of f(n) are 0. An
example is d(n), the number of divisors of n. This multiplicative property
allows such functions to be uniquely defined on the positive integers by
describing the values for positive integer powers of primes. d(pi)=i+1
if i>0; so d(60) = d(22*31*51) =
(2+1)*(1+1)*(1+1) = 12.

Unlike d(n), the sequences described below are a restricted subset of all
multiplicative functions. In all of the cases considered here, f(pi)=pg(i)
for some function g which does not depend on p.

Definition

Multiplicative
with p^i->p^...

Am(n)

Number of solutions to xm == 0 (mod n)

i-ceiling[i/m]

Bm(n)

Largest mth power dividing n

m*floor[i/m]

Cm(n)

mth root of largest mth power dividing n

floor[i/m]

Dm(n)

mth power-free part of n

i-m*floor[i/m]

Em(n)

Smallest number x (x>0) such that n*x is a perfect mth
power (Smarandache mth power complements)

m*ceiling[i/m]-i

Fm(n)

Smallest mth power divisible by n divided by
largest mth power which divides n

m*(ceiling[i/m]-floor[i/m])

Gm(n)

mth root of smallest mth power
divisible by n divided by largest mth power which divides n

Largest mth power-free number dividing n (Smarandache
mth power residues)

min(i,m-1)

Lm(n)

n divided by largest mth power-free number
dividing n

max(0,i-m+1)

Relationships between the functions

Some of these definitions may appear to be similar; for example Dm(n)
and Km(n), but for m>2 all of the functions are distinct (for
some values of n given m). The following relationships follow immediately from
the definitions:

Bm(n)=Cm(n)m

(1)

n=Bm(n)*Dm(n)

(2)

Fm(n)=Dm(n)*Em(n)

(3)

Fm(n)=Gm(n)m

(4)

Hm(n)=n*Em(n)

(5)

Hm(n)=Bm(n)*Fm(n)

(6)

Hm(n)=Jm(n)m

(7)

n=Km(n)*Lm(n)

(8)

These can also be combined to form new relationships; for example from (1),
(4) and (7) we have

Jm(n)=Cm(n)*Gm(n)

(9)

Further relationships can also be derived easily. For example, looking at Am(n),
a number x has the property xm==0 (mod n) if and only if xm
is divisible by n or in other words a multiple of Hm(n), i.e. x is a
multiple of Jm(n). But Jm(n) divides into n, so we have
the pretty result that:

n=Jm(n)*Am(n)

(10)

We could go on to construct a wide variety of further relationships, such as
using (5), (7) and (10) to produce:

nm-1=Em(n)*Am(n)m

(11)

but instead we will note that Cm(n) and Jm(n) are
sufficient to produce all of the functions from Am(n) through to Jm(n):

Am(n)=n/Jm(n)

(12)

Bm(n)=Cm(n)m

Cm(n)=Cm(n)

Dm(n)=n/Cm(n)m

(13)

Em(n)=Jm(n)m/n

(14)

Fm(n)=(Jm(n)/Cm(n))m

(15)

Gm(n)=Jm(n)/Cm(n)

(16)

Hm(n)=Jm(n)m

Jm(n)=Jm(n)

Clearly we could have done something similar by choosing one element each
from two of the sets {A,E,H,J}, {B,C,D}, and {F,G}. The choice of C and J is partly
based on the following attractive property which further deepens the
multiplicative nature of these functions.

If m=a*b then:

Cm(n)=Ca(Cb(n))

(17)

Jm(n)=Ja(Jb(n))

(18)

Duplicate functions when m=2 ...

When m=2, D2(n) is square-free and F2(n) is the
smallest square which is a multiple of D2(n), so

F2(n)=D2(n)2

(19)

Using (3) and (4) we then have:

D2(n)=E2(n)=G2(n)

(20)

and from (13) and (14) we have

n=C2(n)*J2(n)

(21)

so from (10) we get

A2(n)=C2(n)

(22)

... and when m=1

If m=1, all the functions described either produce 1 or n. The analogue of
(20) is still true with

D1(n)=E1(n)=G1(n)=1

(23)

but curiously the analogue of (22) is not, since:

A1(n)=1

(24)

C1(n)=n

(25)

The two remaining functions

All this leaves two slightly different functions to be considered: Km(n)
and Lm(n). They have little connection with the other sequences
except for the fact that since Gm(n) is square-free, and divides Dm(n),
Em(n), Fm(n), and Gm(n), none of which have
any factor which is a higher power than m, we get: